题目链接
题目描述
实现代码
动态规划思想:记dp[i][j]表示text1的前i个字符与text2的前j个字符的最长公共子序列长度,分为两种情况:
- text1[i] == text2[j]时:dp[i][j] = dp[i-1][j-1] + 1
- text1[i] ≠ text2[j]时:dp[i][j] = max(dp[i-1][j], dp[i][j-1])
初始化条件:
- i == 0 时: dp[i][j] = 0
- j == 0 时: dp[i][j] = 0
实现代码:
class Solution {
public int longestCommonSubsequence(String text1, String text2) {
int m = text1.length(), n = text2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 1; i <= m; i++) {
char c1 = text1.charAt(i - 1);
for (int j = 1; j <= n; j++) {
char c2 = text2.charAt(j - 1);
if (c1 == c2) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[m][n];
}
}